Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int num = 0, sgn = 1;
  char ch = getchar();
  while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
  if (ch == '-') sgn = -1, ch = getchar();
  while (ch >= '0' && ch <= '9') num *= 10, num += ch - '0', ch = getchar();
  return num * sgn;
}
int dcmp(double x) {
  if (fabs(x) < 1e-8) return 0;
  return x > 0 ? 1 : -1;
}
struct fence {
  int l, r;
  int type;
} f[400005];
bool cmp(fence A, fence B) { return A.l < B.l; }
int sy, a, b, n, m;
int s[400005];
int q, x, y, lq, rq;
double l, r;
double len, shade;
int find(double x) {
  int l = 1, r = m, mid;
  while (l <= r) {
    mid = (l + r) >> 1;
    if (x > double(f[mid].l))
      l = mid + 1;
    else
      r = mid - 1;
  }
  return r;
}
int main() {
  scanf("%d%d%d", &sy, &a, &b);
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d", &f[i].l, &f[i].r);
    f[i].type = 1;
  }
  sort(f + 1, f + n + 1, cmp);
  m = n;
  for (int i = 1; i < n; i++) {
    f[++m].l = f[i].r;
    f[m].r = f[i + 1].l;
  }
  sort(f + 1, f + m + 1, cmp);
  if (f[m].r < 1000000001) f[++m].l = f[m - 1].r, f[m].r = 1000000001;
  if (f[1].l > 0) f[++m].l = 0, f[m].r = f[1].l;
  sort(f + 1, f + m + 1, cmp);
  for (int i = 1; i <= m; i++) s[i] = s[i - 1] + (f[i].r - f[i].l) * f[i].type;
  scanf("%d", &q);
  while (q--) {
    scanf("%d%d", &x, &y);
    l = double(x) - double(y) / double(y - sy) * double(x - a);
    r = double(x) - double(y) / double(y - sy) * double(x - b);
    len = r - l;
    lq = find(l);
    rq = find(r);
    shade = 0;
    if (f[lq].type == 1) shade += double(f[lq].r) - l;
    if (f[rq].type == 1) shade += r - double(f[rq].l);
    if (rq - lq >= 2) shade += double(s[rq - 1] - s[lq]);
    if (lq == rq) shade = (r - l) * double(f[lq].type);
    printf("%.10lf\n", shade / len * double(b - a));
  }
  return 0;
}